An airline offers a survey for its passengers to complete after every flight. Each time a passenger completes the survey, there is a $2\%$ chance they will win a discounted price on their next flight. Assume that winners are selected at random, and the results of the surveys are independent. Zaylee has numerous trips planned with this airline, and she'll always complete each survey in hopes of winning. Let $N$ be the number of surveys Zaylee completes until she wins for the first time. Find the probability that it takes Zaylee $3$ surveys or less to win for the first time. You may round your answer to the nearest hundredth. $P(N \leq 3)=$
Answer: Without a fancy calculator For each survey: $\begin{aligned} P({\text{win}})&=0.02\\\\ P(\text{not}})&=0.98 \end{aligned}$ If it takes Zaylee $3$ surveys or less to win for the first time, here are the possible sequences of results: win not, win not, not, win We can find the probability of each sequence and add those probabilities together. $\begin{aligned} P({\text{win}}) &= {0.02}\\\\\\ P(\text{not}},{\text{win}}) &= \left(0.98}\right)\left({0.02}\right)\\\\ &=0.0196\\\\\\ P(\text{not}},\text{not}},{\text{win}}) &= \left(0.98}\right)^2\left({0.02}\right)\\\\ &=0.019208\\\\\\ P(N\leq 3) &= 0.02+0.0196+0.019208 \\\\ &=0.058808 \end{aligned}$ [Is there another way?] $P(N \leq 3) \approx 0.06$